0269. Alien Dictionary

https://leetcode.com/problems/alien-dictionary

Description

There is a new alien language that uses the English alphabet. However, the order among the letters is unknown to you.

You are given a list of strings words from the alien language's dictionary, where the strings in words are sorted lexicographically by the rules of this new language.

Return a string of the unique letters in the new alien language sorted in lexicographically increasing order by the new language's rules. If there is no solution, return "". If there are multiple solutions, return any of them.

A string s is lexicographically smaller than a string t if at the first letter where they differ, the letter in s comes before the letter in t in the alien language. If the first min(s.length, t.length) letters are the same, then s is smaller if and only if s.length < t.length.

Example 1:

**Input:** words = ["wrt","wrf","er","ett","rftt"]
**Output:** "wertf"

Example 2:

**Input:** words = ["z","x"]
**Output:** "zx"

Example 3:

**Input:** words = ["z","x","z"]
**Output:** ""
**Explanation:** The order is invalid, so return "".

Constraints:

• 1 <= words.length <= 100

• 1 <= words[i].length <= 100

• words[i] consists of only lowercase English letters.

ac

1. How to intepret as a topological problem?

2. How to build the graph?

class Solution {
    public String alienOrder(String[] words) {
        // edge cases
        if (words == null || words.length == 0)
            return "";

        // map is more robust than array
        Map<Character, Integer> inde = new HashMap<Character, Integer>();
        Map<Character, Set<Character>> g = new HashMap<Character, Set<Character>>();

        // get chars
        for (String w : words) {
            for (int i = 0; i < w.length(); i++) {
                char c = w.charAt(i);
                inde.put(c, 1);
                if(!g.containsKey(c)) g.put(c, new HashSet<Character>());
            }
        }

        // build graph, get indegree
        for (int i = 0; i < words.length - 1; i++) {
            char[] f = words[i].toCharArray();
            char[] s = words[i+1].toCharArray();
            int len = Math.min(f.length, s.length);

            for (int j = 0; j < len; j++) {
                if (f[j] == s[j]) continue;
                if (g.get(f[j]).add(s[j])) { // ignore duplicate edge
                    inde.put(s[j], inde.get(s[j]) + 1);
                }
                break;
            }
        }

        // BFS
        StringBuilder res = new StringBuilder();
        Queue<Character> q = new LinkedList<Character>();
        for (char c : inde.keySet()){
            if (inde.get(c) == 1) q.offer(c);
        }
        while (!q.isEmpty()) {
            char c = q.poll();
            res.append(c);
            for (char next : g.get(c)) {
                inde.put(next, inde.get(next) - 1);
                if (inde.get(next) == 1) q.offer(next);
            }
        }

        // validate
        if (res.length() != inde.size()) {
            return "";
        } else {
            return res.toString();
        }
    }
}
/*
build a graph
indegree
BFS
*/